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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[normed vector spaces|03J5RAKB30]] over $\\R$.\nLet $f \\colon X \\to Y$ be a surjective [[isometry|02HPP5VCZH]], that is, $f$ is onto and $\\norm{f(x) - f(y)} = \\norm{x - y}$ for all $x, y \\in X$.\nThen $f$ is [[affine|0JF0TVYGWT]]: the map $x \\mapsto f(x) - f(0)$ is linear.\n\n## Remarks\n\nThe heart of the proof is that $f$ preserves [[midpoints|0S706JBZBG]]: $f\\left(\\frac{x + y}{2}\\right) = \\frac{f(x) + f(y)}{2}$.\nOne proof describes the midpoint by distances only: it is fixed by every surjective isometry of $X$ that fixes $x$ and $y$.\nThe argument uses the point reflection $z \\mapsto x + y - z$, which swaps $x$ and $y$.\nMidpoint preservation and continuity then give real linearity of $f - f(0)$.\n\nSurjectivity is needed.\nThe map $f \\colon \\R \\to \\R^2$, $f(t) = (t, \\sin t)$, with the norm $\\norm{(a, b)} = \\max(\\abs{a}, \\abs{b})$ on $\\R^2$, is an isometry because $\\abs{\\sin t - \\sin s} \\le \\abs{t - s}$, but it is not affine.\nThe theorem is about real structure only: complex conjugation on $\\C$ is a surjective isometry that is real-linear but not complex-linear.\n\n## Explanations\n\n### L1 human\n\nIf a map between normed spaces keeps all distances and covers the whole target, then it does not bend anything.\nIt is a linear map followed by a shift.\nSo the metric of a normed space already knows its linear structure.\n\n### L2 human\n\nThe theorem shows that two real normed spaces are isometric as metric spaces exactly when they are isometrically isomorphic as normed spaces.\nSo isometric classification of Banach spaces can ignore nonlinear isometries.\nFor a Euclidean space it recovers the fact that every distance-preserving map of $\\R^n$ onto itself is an orthogonal map followed by a translation.\n",
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